1. 1 Law of conservation of energy: In a chemical reaction or physical process, energy can neither be created nor destroyed. Therefore, energy can only be converted from one form to another. Chemical Potential Energy: Energy stored in compounds because of the bonds present in the compound. When a chemical reaction occurs, some bonds are broken and some bonds are made. In many reactions, energy changes are felt as heat.

2. 2 Temperature: a measure of the average kinetic energy of particles in a sample i.e. the faster particles are moving in a sample, the higher the material’s temperature is. Heat: the energy transferred between two materials that are at different temperatures i.e. a substance at a high temperature will transfer heat energy to a substance a low temperature when the two substances come into contact. Heat always flows from hot to cold.

3. 3 Units for Heat: (English system is the calorie, and SI system is the joule) 1 calorie = the heat required to raise the temperature of 1 g of pure water by 1 o C. Nutritional calories are 1000 times larger than a chemical calorie and the units are Cal-notice the capital “C”. 1 Cal = 1000 cal = 1 kcal SI unit for heat is the joule (J). 4.184 J of energy are required to raise the temperature of 1 g of pure water by 1 o C. 1 cal = 4.184 J J = (N)(m) = (kgm/s 2 )(m) = (kg)(m 2 )/(s 2 ) N is Newton and is a unit of force. Force = mass*acceleration = (kg)(m/s 2 ) Force times distance is energy, so a joule is a Newton of force applied over a meter of distance.

4. 4 Convert 245 Cal into cal Convert 245 Cal into kcal Convert 245 Cal into J

5. 5 Specific Heat (c p ) is the amount of heat (J) required to raise the temperature of 1 gram (g) of a substance by 1degree Celsius ( o C) so the units are J/(g o C) The specific heat of water is 4.184 J/(g o C) Notice that specific heat is made up of three units-its units are derived units; Specific heat is not energy, but instead is a physical property of a substance just like density is a physical property. q = (m)(c p )(  T) m is the mass of the substance and the units are usually in grams c p is the specific heat of the substance  T is the change in temperature (  T = T final – T initial ) q is the heat absorbed (+) or released (  ) and usually has units of joules

6. 6 How much heat is required to raise the temperature of 50.0 g of water from 25.0 o C to 35.0 o C? What will the final temperature of a 20.0 g piece of iron be if it absorbs 450.0 J of heat when it was initially 25.0 o C? Specific heat data can be found in figure 1.2 on page 503

7. 7 A simple laboratory calorimeter measures changes in temperature in an insulated environment. A calorimeter is based on the principle that the heat gained by one substance must equal the heat lost by another. For example, the temperature change for water can be measured when a hot piece of metal is placed into a calorimeter filled with water. Since all of the heat required to change the temperature of the water came from the metal, the specific heat of the metal can be determined.

8. 8 A calorimeter contains 125.0 g of water at a temperature of 25.6 o C. A 50.0 g sample of an unknown metal has a temperature of 115.0 o C when it is placed into the calorimeter. The final temperature of the system is 29.3 o C. What is the specific heat of the unknown metal? q H 2 O = (m)(c pH 2 O )(  T) q H 2 O = (125.0 g)(4.184 J/g o C)(29.3 o C – 25.6 o C) = 1935.1 J q gained =  q lost q H 2 O =  q Metal c pMetal = (  q Metal )/(m  T) c pMetal = (  1935.1 J)/(50.0 g)(29.3 o C – 115.0 o C) = 0.451598599 J/g o C = 0.45 J/g o C = 1900 J q cold =  q hot

9. 9 All of the mathematics for changes in energy require that we identify the system being studied. The system is what you are studying (the chemicals and the solvent), the surroundings are everything else in the universe. Therefore, the Universe = System + Surroundings for the purpose of setting up equations. For example: A solution of NaOH is mixed with a solution of HCl. The two solutions are mixed in a beaker and a reaction takes place. The reaction liberates heat. When we study the reaction, the system is the solvent and the chemicals dissolved in it. The beaker and everything outside of the beaker are the surroundings. The aqueous solution contains water (solvent), and a dissolved substance (solute). The system is the solution (solute and solvent). The beaker and everything outside of it are the surroundings.

10. 10 Enthalpy (H): the heat content of a system at constant pressure.  H is the change in enthalpy  H rxn is the change in enthalpy of a reaction  H rxn =  H products   H reactants We are interested in the size and the sign of  H rxn. The size of  H rxn tells us how much heat is involved in a reaction. The sign of  H rxn tells us which direction the heat is flowing during the reaction. If  H rxn is positive (  H rxn > 0), then heat is flowing into the system and  H rxn is said to be endothermic. The Products contain more energy than the Reactants! If  H rxn is negative (  H rxn < 0), then heat is flowing out of the system and  H rxn is said to be exothermic. The Reactants contain more energy than the Products !

11. 11 An exothermic reaction  H rxn =  H products   H reactants  H rxn = negative Notice that the products have less energy than the reactants!

12. 12 An endothermic reaction  H rxn =  H products   H reactants  H rxn = positive Notice that the products have more energy than the reactants!

13. 13 Breaking bonds always requires energy (endothermic process) Making bonds always releases energy (exothermic process) Since a chemical reaction always involves bond breaking and bond making, the energy change for the reaction can be determined by comparing the total energy of all the bonds broken to the total energy of all the bonds being made. If the energy released during bond making is greater than the energy required for bond breaking, a reaction will be exothermic. If the energy released during bond making is less than the energy required for bond breaking, a reaction will be endothermic. Another way of saying this is: if the energy of the products is less than the energy of the reactants, the reaction will be exothermic. If the energy of the products is more than the energy of the reactants, the reaction will be endothermic.

14. 14 Thermochemical Equations A chemical equation that includes the heat of reaction is a thermochemical equation. The heat of reaction may be included in the equation as a reactant or a product, or the heat of reaction can be included as a  H rxn value given along with the equation. If heat is a product, the reaction is exothermic. If heat is a reactant, the reaction is endothermic. 4 Fe (s) + 3 O 2 (g) → 2 Fe 2 O 3 (s) + 1625 kJ NH 4 NO 3 (s) + 27 kJ → NH 4  (aq) + NO 3  (aq) 4 Fe (s) + 3 O 2 (g) → 2 Fe 2 O 3 (s)  H =  1625 kJ NH 4 NO 3 (s) → NH 4  (aq) + NO 3  (aq)  H = 27 kJ

15. 15 4 Fe (s) + 3 O 2 (g) → 2 Fe 2 O 3 (s) + 1625 kJ The heat of reaction can also be used in stoichiometry problems. How many moles of Fe were reacted if 455.7 kJ of heat were released during the above reaction? How many grams of Fe would be required to produce 945.0 kJ of heat using the above reaction? 455.7 kJ ()() 4 mol Fe 1625 kJ = 1.121723 mol Fe= 1.122 mol Fe

16. 16  H rxn for a combustion reaction is renamed  H comb or the heat of combustion. We are most interested in the heat of combustion (  H comb ) because we burn fuels to make energy. If we know the  H comb for a certain type of fuel, we can determine how much fuel will be needed to produce a specific amount of energy. For example: The  H comb for propane (C 3 H 8 ) is  2,219 kJ/mol How much energy would be produced when 50.0 g of propane is burned? CH 4 + 2 O 2 → CO 2 + 2 H 2 O Use: q = (n)(  H) where q is the amount of heat and n is moles

17. 17 Heat changes for phase changes are also given special names. Molar Heat of Vaporization (  H vap ): heat required to vaporize 1 mol of a liquid Molar Heat of Fusion (  H fus ): heat required to melt 1 mol of a solid Heat is required to make this phase change occur, so the processes is endothermic and values for heats of fusion are always positive. Condensation is the opposite process to vaporization, so a heat of condensation (  H cond ) and a heat vaporization have opposite signs but are of equal magnitude.  H cond is exothermic Heat is required to make this phase change occur, so the process is endothermic and values for heats of vaporization are always positive. Solidification is the opposite process to fusion, so a heat of solidification (  H solid ) and a heat fusion have opposite signs but are of equal magnitude.  H solid is exothermic

18. 18  H vap =   H cond  H fus =   H solid  H vap for water is 40.7 kJ/mol and  H fus for water is 6.01 kJ/mol How much energy would be released when 250.0 g of steam condenses at 100 o C? Use: q = (n)(  H) How much energy would be required to melt 50.0 g of ice at 0 o C?

19. 19 What is happening in the places where the graph is flat? A heating curve: heat is added at a constant rate to a solid at low temperature.

20. 20 How much heat is required to convert a 100.0 g block of ice at  25 o C to steam at 115 o C? q total = q 1 + q 2 + q 3 + q 4 + q 5 but q 2 = (n)(  H fus ) and q 4 = (n)(  H vap ) q 1 = (m)(c s )(  T) q 3 = (m)(c l )(  T) q 5 = (m)(c g )(  T) Solid phase Liquid phase Gaseous phase For water: c s = 2.03 J/g o C c l = 4.184 J/g o C c g = 2.01 J/g o C n = moles of substance

21. 21 Hess’s Law: if you add two or more thermochemical equations to produce a final equation, the sum of the enthalpy changes of individual thermochemical equations is equal to the enthalpy change for the overall reaction. S (s) + O 2 (g) → SO 2 (g) + 297 kJ 2 SO 2 (g) + O 2 (g) → 2 SO 3 (g)  H = ? 2 S (s) + 3 O 2 (g) → 2 SO 3 (g) + 792 kJ How can we use Hess’s Law to find the  H rxn for this reaction? Since SO 2 is a reactant in the final equation, reaction 1 must be reversed. Since there are 2 SO 2 in the final equation, reaction 1 must be multiplied by 2. Then add reaction 1 to reaction 2. Reaction 1: Reaction 2:

22. 22 S (s) + O 2 (g) → SO 2 (g) + 297 kJ 2 SO 2 (g) + O 2 (g) → 2 SO 3 (g)  H = ? Target Equation: Reverse this equation SO 2 (g) + 297 kJ → S (s) + O 2 (g) Notice that reversing the equation changes the sign of  H. Multiply by 2 2 SO 2 (g) + 594 kJ → 2 S (s) + 2 O 2 (g) Notice that multiplying the equation by 2 doubles  H. Now add the two equations together. 2 S (s) + 3 O 2 (g) → 2 SO 3 (g) + 792 kJ 2 SO 2 (g) + 594 kJ → 2 S (s) + 2 O 2 (g) 2 SO 2 (g) + O 2 (g) → 2 SO 3 (g) + 792 kJ – 594 kJ198 kJ  H =  198 kJ

23. 23 Rules for using Hess’s Law: Reversing an equation changes the sign of  H (like multiplying by  1) Multiplying an equation by a number requires you to multiply the  H by the same number. 2 SO 2 (g) + O 2 (g) → 2 SO 3 (g) 2 S (s) + 3 O 2 (g) → 2 SO 3 (g) S (s) + O 2 (g) → SO 2 (g)  H =  297 kJ  H =  792 kJ SO 2 (g) → S (s) + O 2 (g)  H = 297 kJ 2SO 2 (g) → 2S (s) + 2O 2 (g)  H = 594 kJ  H =  792 kJ + 594 kJ  H =  198 kJ

24. 24 Standard Enthalpy (Heat) of Formation: HofHof is the change in enthalpy that occurs when one mole of a substance is produced in its standard state from it elements. Even though most compounds can not actually be made directly from their elements, this method allows us to determine the change in enthalpy for reactions using Hess’s Law. The small circle indicates “standard state” which is the normal physical state at 1 atm and 25 o C.  H rxn =  (  H o f (products))   (  H o f (reactants)) The heat of formation of any element in its standard state is 0 kJ.

25. 25 Example: the heat of formation for HF (g) =  273 kJ for SF 6 (g) =  1220. kJ for H 2 S (g) =  21 kJ H 2 S (g) + 4 F 2 (g) → 2 HF (g) + SF 6 (g) What is the heat of reaction for the following reaction?  H o f =  273 kJ ½ H 2 (g) + ½ F 2 (g) → HF (g)  H o f =  1220. kJ S (s) + 3 F 2 (g) → SF 6 (g)  H o f =  21. kJ S (s) + H 2 (g) → H 2 S (g)  H o f =  546 kJ H 2 (g) + F 2 (g) → 2 HF (g)  H o f =  1220. kJ S (s) + 3 F 2 (g) → SF 6 (g)  H o f = 21. kJ H 2 S (g) → S (s) + H 2 (g) H 2 S (g) + 4 F 2 (g) → 2 HF (g) + SF 6 (g)  H o f = – 1745 kJ Applying Hess’s Law to the formation equations gives:

26. 26 Again: the heat of formation for HF (g) =  273 kJ for SF 6 (g) =  1220. kJ for H 2 S (g) =  21 kJ  H rxn =  (  H o f (products))   (  H o f (reactants)) H 2 S (g) + 4 F 2 (g) → 2 HF (g) + SF 6 (g) We can find the same answer using a memorized equation:  (  H o f (products)) = (2(  273 kJ) + (  1220 kJ)) =  1766 kJ  (  H o f (reactants)) = ((  21 kJ) + 4(0 kJ)) =  21 kJ  H o rxn = (  (1766 kJ)  (  21 kJ)) =  1745 kJ Notice that the coefficients of the balanced equation must be included! Now we plug the heat of formation information into the new equation:

27. 27 Entropy (S): a measure of the order (or randomness) of a system. Entropy is an energy term, so its units are joules, but it is not like heat.  S system = S products   S reactants If entropy increases in a reaction, S products > S reactants, so  S > 0 or  S is positive. What does it mean for entropy to be greater in one state than another? Greater entropy means that the system is more disordered than it was before. Compare a new deck of cards to a deck of cards that has been thrown on the floor and then quickly pushed together into a stack. Which has greater entropy?

28. 28 Anything that allows for particles to be arranged in more ways is an indication of greater entropy. Examples: A flask with a pure substance compared to a flask with a mixture. A flask with solid water compared to a flask with liquid water. A flask with 2 moles of chemicals compared to a flask with 4 moles of chemicals. A flask with a gas in it compared to a flask with a solid in it.

29. 29 The entropy of a perfect crystal at 0K is zero. As the temperature is raised, entropy increases.

30. 30 The expansion of an ideal gas into an evacuated bulb. Entropy Increases

31. 31 The amount of motion of the particles effects the number of possible states that the particles can be in.

32. 32 All of these features are possible for reactions.

33. 33 Spontaneous Processes: any process that, once started, will continue on its own without any outside intervention. Example: A ball on the side of a hill will roll to the bottom once it starts rolling. Example: A block of ice will melt if left in a room that is at 25 o C. Example: Steam will condense if left in a room that is at 25 o C. Example: A cup of gasoline will continue to burn in air once it is lit. Non-spontaneous Process: any process that, once started, will not continue on its own without any outside intervention. Example: A cup of water will not freeze if left in a room at 25 o C. Example: A ball will not roll up a hill on its own. Example: A room will not get neater on its own. Start

34. 34 Both  H and  S play a part in determining if a process is spontaneous  G =  H   T  S where T is the Kelvin temperature Change in Free energy (  G) is the concept chemists use to determine whether a process is spontaneous or not. If  G < 0, (negative) a process is spontaneous. If  G > 0, (positive) a process is not spontaneous. If  G = 0, a process is at equilibrium-no longer changing in any direction.

35. 35  G  H   S We can use our understanding of DH and DS for a reaction to make predictions about DG for the reaction. Always Positive (not spontaneous) Positive (endothermic) Negative (becoming more ordered) Negative (exothermic) Positive (becoming more random) Always Negative (spontaneous) Positive (endothermic) Positive (becoming more random) Negative at high T (spontaneous) Negative (exothermic) Negative (becoming more ordered) Negative at low T (spontaneous)

36. 36  G system =  H system  T  S system  G o system =  H o system  T  S o system  G o rxn =  (  G o f (products))   (  G o f (reactants))  H o rxn =  (  H o f (products))   (  H o f (reactants))  S o rxn =  (  S o f (products))   (  S o f (reactants)) If  G o system = 0, then  H o system  T  S o system = 0 and T = (  H o system )/(  S o system ) this is called the crossover temperature Note: S is often given in J/mol·K while H and G are in kJ/mol In order to complete the math correctly, S must be changed to kJ/mol·K. Summary of Important Thermodynamic Equations:  G o rxn =  H o rxn  T  S o rxn